Question

Vectors $\vec{ a }$ and $\vec{ b }$ are inclined at an angle $\theta=120^{\circ}$ If $|\vec{a}|=1,|\vec{b}|=2$ then $\{(\vec{a}+3 \vec{b}) \times(3 \vec{a}-\vec{b})\}^{2}$ is equal to

• A. 225
• B. 275
• C. 325
• D. 300

Answer 1

Answer: (D)

Since, $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos 120^{\circ}$
$=1.2 \cdot\left(-\frac{1}{2}\right)=-1$
$\because\{(\vec{a}+3 \vec{b}) \times(3 \vec{a}-3 \vec{b})\}^{2}$
$=\{3 \vec{a} \times \vec{a}-\vec{a} \times \vec{b}+9 \vec{b} \times \vec{a}-3 \vec{b} \times \vec{b}\}^{2}$
$=[0-\vec{a} \times \vec{b}-9 \vec{a} \times \vec{b}-0]^{2}$
$=100(\vec{a} \times \vec{b})^{2}$
$=100\left\{a^{2} b^{2}-(\vec{a} \times \vec{b})^{2}\right\}$
$=100\{4-1\}=300$